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problem with symbolic factorization with two symbols

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rustem
rustem le 21 Oct 2023
Modifié(e) : Torsten le 21 Oct 2023
The output is not a factorized one. I want this form : (s+s1)*(s+s2)*(s+s3). What is the problem?
syms s K
factor(s^3 + 10*s^2 + (21+K)*s + 4*K, [s, K],'FactorMode','full')
The output is
4*K + 21*s + K*s + 10*s^2 + s^3
  1 commentaire
Dyuman Joshi
Dyuman Joshi le 21 Oct 2023
It is factorized according to the inputs given.
From the documentation - F = factor(x,vars) returns an array of factors F, where vars specifies the variables of interest. All factors not containing a variable in vars are separated into the first entry F(1). The other entries are irreducible factors of x that contain one or more variables from vars.
Note the last sentence.
What is the expected output from the vectorization?

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Réponse acceptée

John D'Errico
John D'Errico le 21 Oct 2023
Modifié(e) : John D'Errico le 21 Oct 2023
That you want to see a simple set of factors is not relevant. You essentially need to obtain the roots of this cubic polynomial in s, where K is a parameter, which would give you that factorization.
syms s K
sroots = solve(s^3 + 10*s^2 + (21+K)*s + 4*K,s,'maxdegree',3)
sroots = 
As you can see, they are fairly messy, and depending on the value of K, they may well be complex even if K is real. This is expected for any cubic polynomial, that we may find exactly two conjugate complex roots.
The factors now are just
prod(s-sroots)
ans = 
We can expand that to show the result recovers your original polynomial.
simplify(expand(ans))
ans = 
Finally, why did not factor understand what you wanted to see? You gave it two variables s and K. How should factor possibly know that you wanted it to do what I just did, essentially, that s is the variable of interest, and K a symbolic parameter in the problem? Software cannot read your mind, at least, not yet.
  7 commentaires
Walter Roberson
Walter Roberson le 21 Oct 2023
Factorization of expressions with real components (or components of unknown reality) traditionally only proceeds to the point where any further factoring would require use of imaginary components.
syms x
factor(x^2 + 1)
ans = 
factor(x^2 - 1)
ans = 
P1 = expand((x-2) * (x+7) * (x-7))
P1 = 
factor(P1)
ans = 
P2 = expand((x-2) * (x+7i) * (x-7i))
P2 = 
factor(P2)
ans = 
P3 = expand((x-2i) * (x+7) * (x-7))
P3 = 
factor(P3)
ans = 
Torsten
Torsten le 21 Oct 2023
Modifié(e) : Torsten le 21 Oct 2023
I cannot see a rule when "factor" works and when if does not work.
syms x
P2 = expand((x-2) * (x+7i) * (x-7i))
P2 = 
factor(P2,x,'FactorMode','complex')
ans = 

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Plus de réponses (1)

Torsten
Torsten le 21 Oct 2023
Here are the factors:
syms s K
eqn = s^3 + 10*s^2 + (21+K)*s + 4*K == 0;
factors = solve(eqn,s,'MaxDegree',3)
factors = 
  2 commentaires
Dyuman Joshi
Dyuman Joshi le 21 Oct 2023
Shouldn't the factors be
s - roots_of_equation
Torsten
Torsten le 21 Oct 2023
Yes, I should have said: -s1, -s2 and -s3 from the factorization (s+s1)*(s+s2)*(s+s3).

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